I am trying prove this:
Let $(X, \tau)$ be a topological space. Suppose $H$ and $H'$ are two connected subespaces such that $ad(H)\cap H'\neq \emptyset$. Prove that $H\cup H'$ is connected.
I have a previous result that says that: if $H$ and $H'$ are two connected subespaces such that $H\cap H'\neq \emptyset$ then $H\cup H'$ is connected.
I have tried to use this previous result, as well as using directly the definition of connected space. But I did not manage to prove it successfully. It would be very helpful if anyone could give me a hint.
Thanks.
I will answer assuming that by $ad(H)$ you mean the closure of $H$.
We will use two facts: the first, which you mention, is that if $H, H'$ are connected and $H \cap H' \neq \emptyset$, then $H\cup H'$ is connected. The second is that if $H$ is connected and $H \subseteq H' \subseteq ad(H)$, then $H'$ is connected.
Now $H \cup H' = (H \cup (ad(H) \cap H')) \cup H'$. By the second fact $H \cup (ad(H) \cap H')$ is connected, and so you conclude by the first fact, using that $(H \cup (ad(H) \cap H') ) \cap H' = ad(H) \cap H' \neq \emptyset$.
Here is the proof of the second fact in case you don't know it. Let $H \subseteq H' \subseteq ad(H)$ and suppose that $H' \subseteq U \sqcup V$, where $U, V$ are open. Since $H$ is connected, $H$ is completely contained in one of the two open sets, say $U$. We claim that $H' \subseteq U$ as well. Indeed, if not, choose $x \in H \cap V$. Since $x \in ad(H)$ and $V$ is a neighbourhood of $x$, we deduce $H \cap V \neq \emptyset$, a contradiction.